Ask question

Ask question

All categories

3 years ago

7

PR and PS are tangent to center Q. Find the measurement of angle Q

1 answer:

iVinArrow [24]3 years ago

8 0

Segment PQ bisects the angle P, which means angle QPR has a measure of 21 degrees (which follows from the fact that triangles PQR and PQS are congruent and PR and PS are tangent to the circle).

This then means angle RQP has measure 90 - 21 = 69 degrees.

Angle Q is twice this, so it has measure 138 degrees.

You might be interested in

K/2 - 1 = 1 solve for k

Nana76 [90]

K=4 would be the answer you’re looking for

6 0

3 years ago

Will give brainliest, *15 points*

olganol [36]

2.4

240u^2 / 100u^2 = 2.4

5 0

2 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

prisoha [69]

<h3>Given:</h3>

h= 19 ft
r= 12 ft

<h3>Note that:</h3>

h= Height
r= Radius

<h3>To find:</h3>

The volume of the given cone.

<h3>Solution:</h3>

$\large\boxed{Formula:V= \frac{1}{3}\pi{r}^{2}h}$

$\large\boxed{\red \pi \red = \red 3 \red . \red 1 \red 4}$

Let's solve!

Substitute the values according to the formula.

$V=\frac{1}{3}×3.14×{12}^{2}×19$

$\large\boxed{V=2863.68 \: {ft}^{3}}$

<u>Therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>volume</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>cone</u><u> </u><u>is</u><u> </u><u>2863.6</u><u>8</u><u> </u><u>cubic</u><u> </u><u>feets</u><u>.</u>

5 0

2 years ago

Using the figure below, select the pair(s) of supplementary angles. (HINT** they may be more than 1 correct answer)

kondor19780726 [428]

Answer:

B, D

Step-by-step explanation:

Supplementary Angles: Add up to 180 degrees

Angle abe and Angle Ebc add up to 180 degrees

Angle ABD and Angle DBC add up to 180 degrees

3 0

3 years ago

Read 2 more answers